Solution procedure

In order to find a function N1 (x) satisfying Equation 5.50 it is necessary to have an expression for the moment distribution M (x) along the beam in question. Here it will be demonstrated how to find a solution for a beam subjected to transverse loading causing a bending moment distribution that can be expressed as a polynomial of order four. Of course higher orders are possible, but are quite rare in practical applications. Note that moment distributions expressed as polynomials are by far the most common type of loading assumed in classic beam bending theory. Besides from this, the beam may have quite general boundary conditions. The moment distribution along the beam may be written as:

where the ai’s are known coefficients for statically determinate beams and for statically indeterminate beams one or more of them must be expressed in terms of some indeterminate quantity, for example an unknown support reaction or bending moment at a built in support. Note that the beam may have to be divided into more than one interval along the x-axis, that is the function representing the moment distribution may in general be different for each interval depending on the actual loading and boundary conditions. Here just one interval having a continuous moment distribution is assumed. Substituting Equation 5.62 into 5.50 gives:

Eq. 5.59

Eq. 5.60

Eq. 5.61

Eq. 5.62

This second order differential equation is linear and has constant coefficients and may be solved using standard mathematical procedures given in most schoolbooks on elementary differential equations. The total solution is the sum of the homogeneous solution and one particular solution: The homogeneous solution is:

where the constants D1 and D2 must be determined with regard to the boundary conditions of the problem. The particular solution must be a fourth order polynomial because the right part of Equation 5.63 is such a polynomial. By substituting a polynomial trial solution and its second order derivative expressed as:

into Equation 5.63 and identifying terms a linear equations system in the unknown bi’s are set up. A symbolic solution to that system expresses the unknown bi’s of the particular solution in the known ai’s, which together with Equation 5.65 are substituted back into Equation 5.64 yielding:

Furthermore, it is usually necessary to have an expression for the deflection function, which is obtained by substituting Equation 5.67 into Equation 5.59 resulting in:

Note that the principal of superposition holds because linear elastic material is assumed, that

Eq. 5.64

Eq. 5.65

Eq. 5.66

Eq. 5.67

solutions to new elementary cases, rather than explicitly solving Equations 5.50 and/or 5.51 possibly with the aid of Equations 5.67 and/or 5.68.

One consequence that immediately follows is that it would not have been necessary to carry all terms including εt through Equations 5.50 to 5.68. If in Equation 5.50 simply setting

M = 0 as if no external bending moment is present, a new elementary case only dealing with εt

is formed. Its solution can then be added to any other case as needed. But now the easiest way to find the solution to this case is to recognise that setting M = 0 is equivalent to setting all ai’s to 0 in Equations 5.67 and 5.68, after which the coefficients D1, D2, D3 and D4 are determined with respect to actual boundary conditions.

It will now be demonstrated how to find the complete solution to one elementary case of practical interest. The case of a simply supported beam subjected to a concentrated load P acting at mid-span is chosen and it is also assumed that εt = 0. The total span is l and the moment distribution has a discontinuous derivative at mid-span, at l /2. It is therefore necessary to divide the beam into two parts one for 0 ≤ x ≤ l /2 and one for l /2 < x ≤ l. But because of symmetry it suffices to find a solution for one of the intervals, preferably 0 ≤ x ≤ l /2. The bending moment is expressed as M(x) = P/2 ∙ x. By comparing to Equation 5.62 we see that a1 = P/2, while a0, a2, a3 and a4 = 0. The coefficients D1, D2, D3 and D4 of Equations 5.67 and 5.68 may now be determined using some adequate boundary conditions. The axial force in each layer as well as the deflection must be zero at x = 0, which may be expressed as N1(0) = 0 and w(0) = 0. These two conditions give without any calculations that

D1 = D4 = 0 in Equations 5.67 and 5.68. At mid-span the axial force N1 and the deflection w have

their maximum and their derivatives must therefore be zero, N'1(l /2) = 0 and w'(l /2) = 0. From Equation 5.59 it follows that D3 = 0. Finally, Equation 5.67 gives:

The total solution is obtained by substituting the results of Equation 5.69 back into Equations 5.67 and 5.68, and after some rearranging of terms and developing of the deflection function (wfca for full composite action) we arrive at:

Eq. 5.69

One can of course find solutions to other problems, being loaded differently and having other boundary conditions, by using the same principal approach. A summarising recipe may look like:

a) Divide if necessary the entire beam into a number of intervals for which the moment distribution is continuous, that is has a continuous first order derivative or in other words can be written as one single function expression. Note also that all cross-sectional con- stants must not change along each chosen interval.

b) For each interval identified under point (a), express the moment distribution as a poly- nomial and identify the ai’s in Equations 5.62, 5.67 and 5.68. Note also that if the beam is statically indeterminate one or more of the ai’s have to be expressed in terms of all the statically indeterminate quantities, such as a reaction force or bending moment. c) Figure out relevant boundary conditions such that Equations 5.67 and 5.68 plus their first

and second order derivatives can be used to find expressions for the coefficients D1, D2,

D3 and D4. Remember that D3 and D4 are often zero due to the structure of Equation 5.59.

One equation is needed for each Di and for each statically indeterminate quantity. Boundary conditions securing continuity of Equations 5.67, 5.68 and their derivatives across interval borders are useful if the beam is subdivided into several intervals. Note also that all Di’s are in general different on each side of an interval border.

d) Set up all chosen boundary conditions as a system of linear algebraic equations and solve it for the unknown Di’s and if present the statically indeterminate quantities. The solution can be either symbolic or numeric. A symbolic solution is more versatile and can be reused over and over again for different input data like dimensions and stiffness properties, while a numeric solution is only valid for exactly the same input data as those used for setting up the equation system. Some useful boundary conditions are: N1 = 0 at a free end whether or not the beam is supported or free to deflect at this end; N'1 = 0 at a position where no slip can occur; w = 0 at a pinned support; w' = 0 at a built in end.

The complete symbolic solutions to a large number of practically useful elementary cases are given in Elsander (1999). Here, the solution to just one more case is given, which is a simply supported beam subjected to uniform loading q. The axial force in part 1 is:

It is also possible to use analogies in order to solve the differential Equation 5.50, because other physical phenomena might be controlled by a differential equation of exactly the same mathematical form. For instance, the equation controlling mixed torsion (that is Saint Venant torsion in combination with warping torsion) has an equivalent mathematical form. Also, the equation controlling deflection of an elastic bar subjected to transverse loading in combination with axial tension is equivalent in shape.

Figure 5.22: Partial composite action in comparison to the upper and lower limits.

By using Equation 5.70 some graphics presenting the basic difference in results between full composite action (FCA), partial composite action (PCA) and no composite action (NCA) have been derived and are presented in Figure 5.22. Parameters used are: l = 2,40 m, P = 5,00 kN,

C = 6,49 × 10-9 _mm-3_{, ω = 9,50 × 10}-4 _mm-1_{, E = 11,0 GPa and Ifca = 3,90 × 10}7 _mm4_{. These}

values are representative for a T-section made of two joists of C24-timber having dimensions 45 × 145 mm (turned edgeways) and 45 × 120 mm (turned flatways). These joists are nailed together using one row of 120 × 3,8 mm smooth grooved nails with a spacing of s = 60 mm. Instantaneous conditions have been assumed.

5.2.1.4 Relations between the slip modulus of mechanical connectors and